let $\{a_{n}\}$ is non-increasing postive sequence;show that

if for $n\ge 2001,na_{n}\le 1$, then for any positive integer numbers $m\ge 2001,x\in R$, we have $$\left|\sum\limits_{k=2001}^{m}a_{k}\sin{(kx)}\right|\le 1+\pi $$

This problem is 2001 china team problem.http://www.doc88.com/p-30356357991.html It is well known that

Fejér-Jackson-Gronwall inequality: $$\sum\limits_{k=1}^{n}\dfrac{\sin{kx}}{k}>0$$ Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson) and we know that have $$\left|\sum\limits_{k=1}^{n}\dfrac{\sin{(kx)}}{k}\right| \le 2\sqrt{\pi}$$ Please check my answer to $\sum\limits_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$

and the simple link: How prove this inequlity $\sum_{k=n}^{2n-3}\frac{|\sin{k}|}{k}<\frac{1}{\sqrt{2}},n\ge 3$ Proving that $\sum_{k=n}^{2n-2} \frac{|\sin k\ |}{k} < 0.7\ln 2$ for $n\ge2$ But for my question How prove it? Thank you

  • $\begingroup$ Are we sure that a_n can be unboundedly negative? Suppose my sequence is such that a_n = -(2+pi), then for the sum m=2001, there is an x such that |a_2001 * sin(2001x)| > 1+pi $\endgroup$
    – Ncat
    Sep 6, 2013 at 2:13
  • $\begingroup$ Oh,sorry, I have edit, Thank you $\endgroup$
    – math110
    Sep 6, 2013 at 4:36

1 Answer 1


If we first rearrange the sum in the following way: $$ \sum_{k=2001}^m a_k\sin(k x) = a_m \sum_{k=2001}^{m}\sin(k x)+(a_{m-1}-a_m)\sum_{k=2001}^{m-1}\sin(k x)+\ldots+(a_{2002}-a_{2003})\sum_{k=2001}^{2002}\sin(k x)+(a_{2001}-a_{2002})\sin(2001x),$$ then compute: $$ \sum_{k=2001}^m \sin(kx) = \frac{\cos\left(\frac{4001x}{2}\right)-\cos\left(\frac{(2m+1)x}{2}\right)}{2\sin(x/2)},$$ we have: $$ \sum_{k=2001}^m a_k\sin(k x) = a_{2001}\cdot\frac{\cos\left(\frac{4001x}{2}\right)-1}{2\sin(x/2)}+a_m\cdot \frac{1-\cos\left(\frac{(2m+1)x}{2}\right)}{2\sin(x/2)}+\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}.$$ Now, since the function $\frac{1-\cos\left(\frac{Nx}{2}\right)}{2\sin(x/2)}$ is non-negative and bounded by $\frac{N}{I_0(1)}<\frac{4N}{5}$, we have: $$ \sum_{k=2001}^m a_k\sin(k x) \leq 2m a_m+\sum_{k=2001}^{m-1}2k(a_k-a_{k+1})\leq 2+2\left(\sum_{k=2001}^{m-1}k a_k-\sum_{k=2001}^{m-1}k a_{k+1}\right),$$ from which: $$\left|\sum_{k=2001}^m a_k\sin(k x)\right| \leq 2+2\log\left(\frac{m}{2001}\right)$$ follows. This is clearly not the theorem's claim, since it depends on $m$. Anyway I hope that my argument can be refined in order to remove that dependence.

A refined inequality is: $$ 0\leq \frac{1-\cos\left(\frac{2n+1}{2}x\right)}{2\sin(\frac{1}{2}x)}\leq\min\left(\frac{(2n+1)^2}{8}x,\frac{1}{\sin(x/2)}\right).$$ By symmetry, we only need to prove the inequality for $x\in[0,\pi]$. If $x=\frac{4}{2h+1}$ with $h\geq m$, then: $$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq \frac{4}{2h+1}\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{(2k+1)^2}{8}\leq \frac{2(m-k)}{2h+1}\leq 1.$$ If $x=\frac{4}{2h+1}$ with $h\in(2001,m-1]$, then: $$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq 1+\frac{a_h-a_m}{2\sin(x/2)}\leq 1+\frac{1}{2h\sin\left(\frac{2}{2h+1}\right)}\leq 2,$$ so the inequality is true in the range $x\in\left[0,\frac{4}{2m+1}\right]$ and we can assume that $x$ is big enough, i.e. $x\geq\frac{4}{4003}$. This gives: $$\sum_{k=2001}^{m-1}(a_k-a_{k+1})\frac{1-\cos\left(\frac{(2k+1)x}{2}\right)}{2\sin(x/2)}\leq\frac{a_{2001}}{2\sin\left(\frac{2}{2003}\right)}\leq\frac{1}{2002 \sin\left(\frac{2}{2003}\right)}\leq 1.$$ Putting all together, we have: $$ \left|\sum_{k=2001}^m a_k \sin(kx)\right|<4+\frac{1}{2001}<1+\pi.$$

  • $\begingroup$ It's nice, Thank you very much +1 $\endgroup$
    – math110
    Oct 13, 2013 at 12:59

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